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Classical first-law product leaves entropy-temperature calibration free — Examples

1 · Prerequisites

Classical first-law product leaves entropy-temperature calibration free

2 · Summary

The first example instantiates two distinct positive calibrations on a smooth one-dimensional state family and checks that both reproduce the same energy differential. The counterexample uses two explicit density matrices with the same fixed positive real scalar label and zero energy but distinct finite spectral entropies; it makes no claim about a coupled gravitational model.

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

None yet.

5 · Examples, experiments and counterexamples

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

Two entropy-temperature calibrations preserve the same energy differential

Example

Use the abstract state family $X = (0, \infty)$ with coordinate $r$ , and define $E (r) = r$ , $A (r) = r$ , $κ (r) = 1$ , and $C = 1$ . These are formal real-valued functions with no physical units assigned. Then $d E = C κ d A = d r$ . The choices $b = 1$ and $b = 2$ , with offset $s_{*} = 0$ , give $(T_{1}, S_{1, 0}) = (1, r)$ and $(T_{2}, S_{2, 0}) = (2, r / 2)$ .

Facts & Assumptions

Given: The stated smooth state family and the calibration-freedom lemma.

[A1]

On $X = (0, \infty)$ , $E (r) = A (r) = r$ , $κ (r) = 1$ , and $C = 1$ .

[F1]

Classical first-law product leaves entropy-temperature calibration free: For every constant $b > 0$ and offset $s_{*} \in R$ , set $T_{b} : = b κ$ and $S_{b, s_{*}} : = (C / b) A + s_{*}$ . Then $T_{b} d S_{b, s_{*}} = d E$ . Distinct positive $b$ give distinct functions $T_{b}$ , and changing $s_{*}$ changes $S_{b, s_{*}}$ without changing its differential.

Verification

Given: [A1] and [F1].

1.1givenA1algebra

The family $X = (0, \infty)$ is a nonempty connected smooth manifold, the displayed functions are smooth, $C = 1 > 0$ , and $κ (r) = 1 > 0$ . Direct differentiation gives $d E = d r = C κ d A$ , so the lemma's hypotheses hold.

2.1F1A1step 1.1algebra∎

Apply [F1] with $(b, s_{*}) = (1, 0)$ and $(2, 0)$ . The resulting functions are $(T_{1}, S_{1, 0}) = (1, r)$ and $(T_{2}, S_{2, 0}) = (2, r / 2)$ ; hence $T_{1} d S_{1, 0} = 1 \cdot d r = d r$ and $T_{2} d S_{2, 0} = 2 \cdot (1 / 2) d r = d r = d E$ , although the temperatures and entropy gradients differ.

CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

A fixed scalar label and zero energy do not determine finite spectral entropy

Statement refuted

On the complex Hilbert space $H = C^{2}$ , a density matrix means a positive semidefinite $2 \times 2$ matrix $ρ$ with $Tr ρ = 1$ . Let the Hamiltonian be the zero self-adjoint matrix $H = 0$ , define its energy by $E_{H} (ρ) = Tr (ρ H)$ , fix one positive real scalar label $a > 0$ , and fix a positive scalar $k_{B} > 0$ . Define the finite spectral entropy by $S (ρ) = - k_{B} \sum_{λ \in spec (ρ)} λ \log λ,$ where eigenvalues are counted with multiplicity, $\log$ is the natural logarithm, and $0 \log 0 : = 0$ . The statement that the fixed data $(a, E_{H} (ρ)) = (a, 0)$ determine one value of $S (ρ)$ for every density matrix $ρ$ on $H$ is false.

The scalar $a$ is held fixed as a mathematical label; $H$ is only the displayed finite matrix. No physical units, dynamics, horizon state, or coupling to geometry is part of this construction.

Facts & Assumptions

Given: The definitions in the Statement, with fixed $a > 0$ and $k_{B} > 0$ .

[A1]

The space is $H = C^{2}$ , $H = 0$ , $a > 0$ is fixed, and $k_{B} > 0$ is fixed.

[A2]

For a density matrix $ρ$ , $E_{H} (ρ) = Tr (ρ H)$ and $S (ρ) = - k_{B} \sum_{λ \in spec (ρ)} λ \log λ$ , with eigenvalues counted with multiplicity and $0 \log 0 = 0$ .

Counterexample

Given: Use the two explicit diagonal matrices below with the same $a$ , $H$ , and $k_{B}$ .

1.1givenA1constructalgebra

Let $ρ_{1} = diag (1, 0)$ and $ρ_{2} = diag (1 / 2, 1 / 2)$ . Both are positive semidefinite and have trace one, so they are density matrices; since $H = 0$ , both have energy $E_{H} (ρ_{i}) = Tr (ρ_{i} H) = 0$ , and both share the same fixed datum $a$ .

2.1givenA2step 1.1algebra∎

The eigenvalues of $ρ_{1}$ are $1, 0$ , so [A2] gives $S (ρ_{1}) = - k_{B} (1 \log 1 + 0 \log 0) = 0$ . The eigenvalues of $ρ_{2}$ are $1 / 2, 1 / 2$ , so $S (ρ_{2}) = - k_{B} \log (1 / 2) = k_{B} \log 2 > 0$ . Thus the same scalar-label and energy data occur with different finite entropies.

Remarks

This finite-dimensional matrix counterexample does not assert a coupled gravitational solution, backreaction, a horizon state, or a continuum entropy construction. It only shows that the displayed scalar label and energy value do not select a unique finite spectral entropy.

Classical first-law product leaves entropy-temperature calibration free — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

5 · Examples, experiments and counterexamples

Two entropy-temperature calibrations preserve the same energy differential

Example

Facts & Assumptions

Verification

A fixed scalar label and zero energy do not determine finite spectral entropy

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Sources